PEA Analytical Skills lpu mte practice sheet. All topics all unites.

1. /
Percentage and Pro t loss, Average and S.I. & C.I
Test Summary
No. of Sections: 1
No. of Questions: 45
Total Duration: 60 min
Section 1 - Quants
Section Summary
No. of Questions: 45
Duration: 60 min
Additional Instructions:
None
Q1. If Sarvesh’s income is 40% more than Rupesh’s, how much is Rupesh’s income less than Sarvesh’s?
26.30%
27.50%
28.57%
27.70%
Q2. The price of the sugar falls by 10%. How many quintals can be bought for the same money which was suf cient to buy 18 quintals
at the higher price?
12
15
20
18
Q3. Ten percent of the inhabitants of a town have died of plague, panic set in, during which 25% of the remaining inhabitants left the
town if the population was then reduced to 6750, then what was the originally?
9500
9000
8750
10000
Q4. What percent of 2/9 is 3/45?

2. /
20.5%
10%
20%
30%
Q5. The population of a town is 32000. It increases at the rate of 15% annually. What will be the population 2 years hence?
32000
42320
42230
43220
Q6. If x is 20 percent more than y and y is 50 percent less than z, then x is what percent of z?
25%
75%
100%
60%
Q7. In a examination, 1800 candidates were boys and 2200 were girls. If 42% of boys and 48% of girls passed the examination, then
what is the overall percentage of candidates who failed?
45.3%
58%
52%
54.7%
Q8. The value of a house depreciates each year, by 3/4 of its initial value at the beginning of the year. If the initial value of the scooter is
Rs. 40, 000. What will be the value at the end of 3 yrs?
Rs. 19000
Rs. 16875

3. /
Rs. 17525
Rs. 18000
Q9. A man spends 70% of his income. His income increases by 24% and his expenditure also increases by 15%. Find the percentage
increase in his savings.
35%
24%
45%
55%
Q10. If price of a book is rst decreased by 25% and then increased by 20%, the net change in the price of the book will be:
10% decrease
5% increase
5% decrease
none of these
Q11. Two numbers are 50% and 75% lesser than a third number. By how much percent is the second number to be enhanced to make it
equal to the rst number?
50
100
45
25
Q12. In an examination, a student gets 20% of total marks and fails by 30 marks. Another student gets 32% of total marks which is more
than the minimum pass marks by 42 marks. The pass percentage is
25
50
75
10
Q13.

4. /
A shopkeeper sold goods for Rs. 2400 and made a pro t of 25% in the process. Find his pro t per cent if he had sold his goods for
Rs. 2040.
6.25%
7 %
6.20 %
6.5 %
Q14. A shopkeeper sells two items at the same price. If he sells one of them at a pro t of 10% and the other at a loss of 10%, nd the
percentage pro t/loss.
1% loss
1.5% pro t
2% loss
2% pro t
Q15. If the cost price of 20 articles is equal to the selling price of 15 articles. What is the percentage of pro t or loss that the merchant
makes?
33.33 % pro t
25 % pro t
33.33% loss
40.33% loss
Q16. A merchant makes a pro t of 20% even after allowing a discount of 20% on the marked price. What should be the marked price if
the cost price of the article is Rs.300?
Rs.450
Rs.250
Rs.380
Rs.590
Q17. The cost price of 50 mangoes is equal to the selling price of 40 mangoes. Find the percentage pro t?

5. /
20%
25%
30%
None of these
Q18. A dozen pairs of gloves quoted at Rs.80 are available at a discount of 10%. Find how many pairs of gloves can be bought for Rs.24.
4
5
6
8
Q19. A cellular phone when sold for Rs.4600 fetches a pro t of 15%. Find the cost price of the cellular phone.
Rs.4300
Rs.4150
Rs.4000
Rs.4500
Q20. Pro t after selling a product for Rs. 850 is the same as the loss after selling it for Rs.710. What is the cost of the product?
620
680
700
780
Q21. In order that there may be a pro t of 15% after allowing a discount of 10% on the marked price, The marked price of an book has to
be increased by %?
26.2%
27.7%

6. /
28.4%
32.2%
Q22. Two successive discounts of 20% and 10% is given for an Article. Find the overall Discount % ?
30 %
28 %
32 %
25 %
Q23. An article sold for Rs.96.If percentage of pro t was numerically equal to the cost price ,the cost price of the article is
Rs.80
Rs.60
Rs.90
Rs.100
Q24. A dealer offers three successive discounts of 50%, 20% and 10% on an article. What is the single effective discount rate?
60%
62%
63%
64%
Q25. In the rst 20 overs of a cricket game, the run rate was only 4.5. What should be the run rate in the remaining 30 overs to reach a
target of 300 runs ?
8
6
5
7
Q26. The average age of a group of 8 persons is 40 years. A person aged 60 years leaves the group and another person aged 44 years
joins the group. The new average age of the group is

7. /
44 years
38 years
45 years
40 years
Q27. The average age of a class of 30 students and a teacher reduces by 0.5 years if we exclude the teacher. If the initial average is 14
years, nd the age of the class teacher
22 years
25 years
30 years
29 years
Q28. The average number of visitors of a library in the rst 4 days of a week was 58. The average for the 2nd, 3rd, 4th and 5th days was
60.If the number of visitors on the 1st and 5th days were in the ratio 7:8 then what is the number of visitors on the 5th day of the
library?
18
64
58
56
Q29. The average age of a family of 4 members 3 years ago is 21 years. A baby is born and now the average age of the family is same
as before. Find the age of baby.
8 yrs
9 yrs
10 yrs
11 yrs
Q30. In a hostel there are 30 students and if the number of students increased by 5 then the expense is increased by 40 per day. But the
average expenditure diminishes by 3. Find the original expenditure.

8. /
810
870
910
950
Q31. A tabulator while calculating the average marks of 100 students of an examination, by mistake enters a student’s marks as 68,
instead of 86 and obtained the average as 58; the actual average is
58
58.18
58.8
60
Q32. Find the average of rst 20 multiples of 7?
72.5
73.5
71.5
72.75
Q33. What is the average of the sum of rst 20 odd natural numbers?
20
15
164
21
Q34. A sum of money becomes 4 times at simple interest in 10 years. What is the rate of interest?
10%
20%
30%

9. /
40%
Q35. If a certain sum is doubled in 8 years on simple interest then in how many years will it be four times?
24
16
20
18
Q36. If the compound interest on a certain sum for 2 years is Rs.21. What could be the simple interest?
Rs.20
Rs.16
Rs.18
Rs.2.5
Q37. In what time will Rs.390625 amount to Rs 456976 at 4% compound interest?
3 years
4 years
5 years
None of these
Q38. Find the simple interest and compound interest of Rs 15000 at 20% rate of interest after 3 years.
9000, 11000
8000, 11920
9000, 10920
6000, 9000
Q39. A sum becomes triple in 6 years at S.I. The same sum will become 19 times in how many years?

10. /
50 years
48 years
54 years
57 years
Q40. A sum of Rs 343 becomes 512 in 3 years at C.I. Find the rate of interest
14 (2/7) %
12.5 %
8 (2/3) %
16 (2/3)
Q41. If the ratio of difference between CI and SI for 3 years and 2 years is 31:10, then nd the Rate of Interest.
11.11%
10%
20%
25%
Q42. A sum of money in 3 years becomes 1736 and in 7 years it becomes Rs.1988. What is the principal sum where simple rate of
interest is to be charged?
1500
1547
1475
1447
Q43. The difference between CI and SI on a sum of money lent for 2 years at 10% is Rs.40. The sum is:
1600
2000
4000

11. /
None of these
Q44. A sum of Rs. 12,000 deposited at compound interest becomes double after 5 years. After 20 years, it will become :
96000
120000
124000
192000
Q45. A sum of money placed at compound interest doubles itself in 3 years. In how many years will it amount to 8 times itself?
9 years
8 years
27 years
7 years

12. /
Answer Key & Solution
Section 1 - Quants
Q1
Solution
Let Rupesh's income be 100 and Sarvesh's income be 140
Rupesh's income percentage less = (40/140) x 100 = 28.57%
28.57%
Q2
Solution
Let the price per quintal price be x.
If money is m we have m/x =18 and
We have to nd m/(0.9 x) = 18/0.9 = 20quintals
20
Q3
Solution
Let the total Population = X
Now (90/100)(75/100)X= 6750
X= 10000
10000
Q4
Solution
Required % = × 100 = 30%.
30%
Q5
Solution
Based on formula => 32000 [1 + (15/100)]2 = 42320.
42320
Q6
Solution
Lets take values of x,y,z which can satisfy given equations.
60%

13. /
Let Z= 200, so Y becomes 100 ( since Y is 50% less than Z),
X becomes 120 ( since 20% more than Y)
now req is (X/Z) * 100 = (120/200) *100 = 60 %
Q7
Solution
The total number of candidates = 1800 + 2200 = 4000.
Number of boys failed = (100 – 42)% of 1800 = 1044.
Number of girls failed = (100 – 48)% of 2200 = 1144.
Number of candidates failed = 1044 + 1144 = 2188.
Overall percentage of failure =2188/4000 × 100 = 54.7%.
54.7%
Q8
Solution
his is the question of succession depreciation. the starting amount = Rs. 40000 This reduces by 3/4 th of its initial value every year = (40,
000) * (3/4)^3 = 16875
Rs. 16875
Q9
Solution
Let the income of the man be Rs 100
He used to spend= 70 % of his income = Rs 70
And he used to save = Rs 100 – Rs 70 = Rs 30 = 30% of his income.
Now, his income increased by 24%.
∴ New Income = Rs 100 + 24/100 ×Rs 100 = Rs 100 + Rs 24 = Rs 124.
45%

14. /
His expenditure increases by 15%
∴ New Expenditure = Rs 70 + 15/100 ×Rs 70 = Rs 70 + Rs 10.50 = Rs 80.50
∴ His New Savings = Rs 124 – Rs 80.50 = Rs 43.50
∴ Increase in Savings = Rs 43.50 – Rs 30 = Rs 13.50
∴ % Increase in Savings = 13.5 / 30 * 100 = 45%
Q10
Solution
x1+x2+x1x2/100
-25+20-25*20/100
20-30
-10
10% decrease
Q11
Solution
Let the third number = x
⇒ The rst number = x - 50% of x = x/2
⇒ The second number = x - 75% of x = x/4
⇒ The percent by which second number has to be enhanced to make it equal to the rst number = [(x/2 - x/4) / (x/4)] × 100 = 100%
(It can also be seen that x/4 is just half of x/2 so it has to be multiplied twice to become the second number which is 100%)
100
Q12
Solution
Let the total marks be X
Then, marks obtained by student 1 = 0.2X
25

15. /
Since, student 1 fails by 30 marks after getting 20% marks
∴ According to student1, Passing marks = 0.2X + 30
Similarly, Marks obtained by student 2 = 0.32X
And according to student 2, passing marks = 0.32X - 42;
Equating eq. 1 and 2, we get
0.2X + 30 = 0.32X - 42
⇒ X = 600
∴ Passing marks = (0.2 × 600) + 30 = 150
Passing marks % = (150/600) *100= 25%
Q13
Solution
125% = 2400
100% = 1920
% gain obtained for SP of 2040 = (2040-1920)*100/1920
= (120/1920) * 100 = 6.25%
6.25%
Q14
Solution
The result will always be a loss of [ 10/10]^2%. Hence, the answer here is [10/10]^2% = 1% loss.
1% loss
Q15
Solution
Cost Price of 20 articles= Selling Price of 15 articles
20 CP= 15 SP
CP/SP= 15/20= 3/4
Pro t %= (P/CP)*100= (1/3)*100= 33.33%
33.33 % pro t
Q16
Solution
CP of the article = 300.
Rs.450

16. /
SP of the article = 360. (which means 20 % pro t)
Let marked price be x.
80 % of x = 360
x = 360*(100/80)
x = 450
Q17
Solution
50 cp =40 sp
cp/sp = 40/50
=> Pro t = (50-40)/40
= 10/40 x 100
25% pro t
25%
Q18
Solution
Given MP =80 Rs
10% dis gives SP =72
So, 12 pairs of glove =72 Rs
1 pair of gloves 6 Rs
So, for 24 Rs we can buy 4 pairs
4
Q19
Solution
15% pro t.
CP 4600
(1.15 ) CP =4600
CP =4600/1.15
=4000
Rs.4000

17. /
Q20
Solution
Let the CP be Rs. x.
Therefore,
when sold at Rs. 850, Pro t=850-x
when sold at Rs. 710, Loss=x-710
2x=1560
Rs.780
780
Q21
Solution
From the given data,
Let CP, SP and MP be the Cost price, Selling price and Marked Price.
P%= 15%
D%= 10%
SP= 0.9 MP
SP= 1.15 CP
0.9 MP= 1.15 CP
MP= 1.277777 CP
Mark-Up %= 27.77%
27.7%
Q22
Solution
Overall Discount= d1 + d2 - {(d1*d2)/100}
= 20 + 10 - (20*10/100)
= 30 - 2= 28%
28 %
Q23
Solution
Selling price = 96
Cost price = C
Rs.60

18. /
Pro t percentage = Cost Price = C
Pro t percentage = (Selling Price - Cost Price) / Cost Price
C/ 100 = (96-C)/C
C = 60
Q24
Solution
Effective discount = 100 – (0.5 × 0.8 × 0.9 × 100) = 64%.
64%
Q25
Solution
Run scored in rst 20 overs = 20 x 4.5 = 90 runs.
210 more runs has to be scored to reach target
210/30 = 7 runs/over.
7
Q26
Solution
Total age of 8 persons = 8 × 40 = 320 years.
Total age of the present group of 8 persons = 320 – 60 + 44 = 304 years. .·.
The average = 304/8 = 38 years.
38 years
Q27
Solution
Normal process:
Age of teacher = Total age of (students + teacher)- Total age of students
= (31 × 14) - (30 × 13.5)
= 434 - 405
= 29 years
29 years
Q28
Solution
64

19. /
Average of rst four days= (D1+D2+D3+D4)/4= 58
Total of four days= 58*4= 232----------------------->1
Average of next four days= (D2+D3+D4+D5)/4= 60
Total of next four days= 60*4= 240----------------->2
Now subract equation 1 from equation 2, we get D5-D1= 240-232= 8
also, D1:D5= 7:8
8x-7x=x= 8.
D5= 8x= 64
Q29
Solution
Average age of 4 members, 3 years ago= 21
Average age of 4 members now= 21+3= 24
Total= 24*4= 96 years
Average including baby= 21
Total = 21*5= 105
Age of the Baby= 105-96= 9 years
9 yrs
Q30
Solution
Let average expenditure be P.
(30*P + 40)/35 = P-3
P = 29. So expenditure = 29*30 = 870
870
Q31
Solution
Let the sum of the marks of the remaining 99 students be N.
⇒N+68/100=58
⇒N+68=5800
⇒ N = 5800 – 68 = 5732
Hence, calculating the original average, we have = 5732+86/100
=58.18
58.18

20. /
Q32
Solution
Average of rst 20 multiples of 7= (7+14+21+......+140)/20= 7(1+2+3.....+20)/20= (7*20*21)/(2*20)= 73.5
Average of an A.P is average of rst and last term i.e(7+140)/2= 73.5
73.5
Q33
Solution
Note the average of n odd numbers starting at 1 is equal to n.
1 Odd number: 1 (avg=1)
2 Odd numbers: 1 3 (avg=2)
3 Odd numbers 1 3 5 (avg=3)
4 Odd numbers 1 3 5 7 (avg=4)
5 Odd numbers 1 3 5 7 9 (avg=5)
hence average of rst 20 odd natural number is 20
20
Q34
Solution
Let the principal be x
Now we are given that a sum of money becomes 4 times
Amount = 4x
So, Simple interest = 4x-x = 3x
Time = 10 years
Formula:
30%

21. /
Hence the rate of interest is 30%
Q35
Solution
Let the principle be 100.
Then amount will be 200.
Interest= 100
Time = 8 yrs
R=?
I=PRT/100
100=100*R*8/100
100*100/100*8=R
100/8=R
NOW SUM SHOULD BE 4 TIMES MEANS 400
MEANS INTEREST ON 100 IS 300.
I=PRT/100
300=100*100/8*T/100
300*8/100=T
24= T
24
Q36
Solution
Let the principal be 100
[1 + R]2 = 121
1 + R = 11
R = 10%
SI = (100 * 2 * 10) / 100
= 20
Rs.20
Q37
Solution
4 years

22. /
=> P(1 + r/100)n = A => 390625(1 + 4/100)n = 456976 => (1 + 4/100)n = 456976/390625.
=> (26/25)n = (26/25)4 => n = 4 => the required time is 4 years.
Q38
Solution
P=15000
R=20%
T=3 years
SI = 15000x20x3/100
SI = 9000
Amount(in C.I) = 15000(1 + 20/100)3
A = 25920 =
CI = A-P= 25920- 15000= 10920
9000, 10920
Q39
Solution
When sum triples , Interest doubles in 6 years
When sum becomes 19 times, Interest become 18 times
2I= 6 years
18I= (6*9= 54 years)
Since we have given that
Number of years = 6
Let Principal be 'P'.
Let amount be '3P'
So, Simple interest = A- P = 3P - P = 2P
So, it becomes,
54 years

23. /
Amount = 19 P
Principal = P
Simple interest = A-P = 19P-P = 18P
So, it becomes,
Hence, it will take 54 years to get the sum 19 times.
Q40
Solution
Sum=353; Amount=512
as many year, put that many root i.e
cuberoot(343): cuberoot(512)
7:8
rate=(8-7)/7 *100 =14 (2/7)%
14 (2/7) %
Q41
Solution
Difference for 3 years= Pr2/1002(3 + r/100)---------1
Difference for 2 years= Pr2/1002----------------------------2
Dividing 1 by 2 we get,
(3+r/100)= 31/10
Solving we get r= 10%
10%
Q42
Solution
1547

24. /
Amount in S.I for 3 years= Rs. 1736
Amount in S.I for 7 years= Rs. 1988
Interest for 4 years= 252
Interest for 3 years= (252/4)*3= 189
P= Rs. 1736- 189= Rs. 1547
Q43
Solution
D=P(r/100)2
40 = P(10/100)2
P=4000
4000
Q44
Solution
192000
Q45
Solution
If the money doubles in 3 years,then it will become 4 times in 6 years and 8 times in 9 years.
9 years